Trigonometric formulae, equations and inequalities

Unit 1 Trigonometric formulae, equations and inequalities

Key unit competence: Use trigonometric formulae, equations and inequalities in solving related problems.

Key ideas

• Periods: $\sin x$ and $\cos x$ have period $2\pi$; $\tan x$ has period $\pi$; $\sin (kx)$ has period $\frac{2\pi }{|k|}$.
• Addition formulae: $\sin (a\pm b)=\sin a\cos b\pm \cos a\sin b$; $\cos (a\pm b)=\cos a\cos b\mp \sin a\sin b$; $\tan (a\pm b)=\frac{\tan a\pm \tan b}{1\mp \tan a\tan b}$.
• Double angle: $\sin 2a=2\sin a\cos a$; $\cos 2a={\cos }^{2}a-{\sin }^{2}a=2{\cos }^{2}a-1=1-2{\sin }^{2}a$; $\tan 2a=\frac{2\tan a}{1-{\tan }^{2}a}$.
• Half angle: ${\cos }^2\frac{a}{2}=\frac{1+\cos a}{2}$, ${\sin }^2\frac{a}{2}=\frac{1-\cos a}{2}$; with $t=\tan \frac{a}{2}$: $\sin a=\frac{2t}{1+t^2}$, $\cos a=\frac{1-t^2}{1+t^2}$.
• Product to sum: $2\sin a\cos b=\sin (a+b)+\sin (a-b)$, $2\cos a\cos b=\cos (a+b)+\cos (a-b)$, $2\sin a\sin b=\cos (a-b)-\cos (a+b)$. Sum to product: $\sin p+\sin q=2\sin \frac{p+q}{2}\cos \frac{p-q}{2}$, $\cos p+\cos q=2\cos \frac{p+q}{2}\cos \frac{p-q}{2}$.
• Basic equations: $\sin x=\sin \alpha ⟺x=\alpha +2k\pi$ or $x=\pi -\alpha +2k\pi$; $\cos x=\cos \alpha ⟺x=\pm \alpha +2k\pi$; $\tan x=\tan \alpha ⟺x=\alpha +k\pi$ ($k\in \mathbb{Z}$).
• $a\sin x+b\cos x=c$: write $a\sin x+b\cos x=r\sin (x+\phi )$ with $r=\sqrt{a^2+b^2}$, $\cos \phi =\frac{a}{r}$, $\sin \phi =\frac{b}{r}$. Solutions exist only if $|c|\le r$.
• Inequalities are solved on one period with the unit circle or graph, then extended by the period.

Formulas to know

Common misconceptions

• "$\sin (a+b)=\sin a+\sin b$." No: use the addition formula.
• "$\sin x=\frac{1}{2}$ has the single solution $x=\frac{\pi }{6}$." No: $x=\frac{\pi }{6}+2k\pi$ or $x=\frac{5\pi }{6}+2k\pi$.
• "$\sin 2x=2\sin x$." No: $\sin 2x=2\sin x\cos x$.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Find the exact value of $\tan 15^{\circ }$.

   Show answer

   $\tan (45^{\circ }-30^{\circ })=\frac{1-\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}=\frac{3-\sqrt{3}}{3+\sqrt{3}}=2-\sqrt{3}$.

   Common mistake: Using $\tan 45^{\circ }-\tan 30^{\circ }$.

2. Show that $\frac{1-\cos 2x}{\sin 2x}=\tan x$.

   Show answer

   $\frac{2{\sin }^{2}x}{2\sin x\cos x}=\frac{\sin x}{\cos x}$.

   Common mistake: Using $\cos 2x=1+2{\sin }^{2}x$.

3. Prove that ${\cos }^{4}x-{\sin }^{4}x=\cos 2x$.

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   $({\cos }^{2}x-{\sin }^{2}x)({\cos }^{2}x+{\sin }^{2}x)=\cos 2x\times 1$.

   Common mistake: Expanding $({\cos }^{2}x-{\sin }^{2}x{)}^2$ instead of factorising.

4. Solve $2\cos x+1<0$ for $x\in [0,2\pi [$.

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   $\cos x<-\frac{1}{2}$: $\frac{2\pi }{3}<x<\frac{4\pi }{3}$.

   Common mistake: Including the endpoints.

5. Solve $\cos 2x=\cos x$ for $x\in [0,2\pi [$.

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   $2{\cos }^{2}x-\cos x-1=0$, $(2\cos x+1)(\cos x-1)=0$: $x=0$, $\frac{2\pi }{3}$, $\frac{4\pi }{3}$.

   Common mistake: Using $\cos 2x=2\cos x$.

6. Transform $2\sin 5x\cos 3x$ into a sum.

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   $\sin 8x+\sin 2x$.

   Common mistake: Getting $\sin 15x$.

7. Express $\sin x-\sqrt{3}\cos x$ in the form $r\sin (x-\phi )$ and give its maximum.

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   $r=2$, $\cos \phi =\frac{1}{2}$, $\sin \phi =\frac{\sqrt{3}}{2}$: $2\sin \left(x-\frac{\pi }{3}\right)$; maximum 2.

   Common mistake: Taking $r=1+\sqrt{3}$.

8. Write the general solution of $\sin 3x=\frac{\sqrt{2}}{2}$.

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   $3x=\frac{\pi }{4}+2k\pi$ or $3x=\frac{3\pi }{4}+2k\pi$: $x=\frac{\pi }{12}+\frac{2k\pi }{3}$ or $x=\frac{\pi }{4}+\frac{2k\pi }{3}$.

   Common mistake: Dividing only $\frac{\pi }{4}$ by 3, not the period.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find the exact value of $\tan 15^{\circ }$.	a. $\frac{\sqrt{6}+\sqrt{2}}{4}$
   2. Find the exact value of $\sin 75^{\circ }\cos 15^{\circ }$.	b. $\frac{2+\sqrt{3}}{4}$
   3. Find the exact value of ${\cos }^2\frac{\pi }{8}$.	c. $\frac{2+\sqrt{2}}{4}$
   4. Find the exact value of $\sin 105^{\circ }$.	d. $2-\sqrt{3}$

   Show answer

   1 → d ($2-\sqrt{3}$) · 2 → b ($\frac{2+\sqrt{3}}{4}$) · 3 → c ($\frac{2+\sqrt{2}}{4}$) · 4 → a ($\frac{\sqrt{6}+\sqrt{2}}{4}$)

   • Step 1. $\tan (45^{\circ }-30^{\circ })=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$.
     Step 2. Multiply by $\frac{\sqrt{3}-1}{\sqrt{3}-1}$: $\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$.
   • Step 1. $\sin a\cos b=\frac{1}{2}\left[\sin (a+b)+\sin (a-b)\right]$.
     Step 2. $\frac{1}{2}\left[\sin 90^{\circ }+\sin 60^{\circ }\right]=\frac{1}{2}\left(1+\frac{\sqrt{3}}{2}\right)=\frac{2+\sqrt{3}}{4}$.
   • Step 1. Half-angle: ${\cos }^2\frac{\pi }{8}=\frac{1+\cos \frac{\pi }{4}}{2}$.
     Step 2. $\frac{1+\frac{\sqrt{2}}{2}}{2}=\frac{2+\sqrt{2}}{4}$.
   • Step 1. $105^{\circ }=60^{\circ }+45^{\circ }$.
     Step 2. $\sin 60^{\circ }\cos 45^{\circ }+\cos 60^{\circ }\sin 45^{\circ }=\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2}}{2}+\frac{1}{2}\cdot \frac{\sqrt{2}}{2}$.
     Step 3. $=\frac{\sqrt{6}+\sqrt{2}}{4}$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $\frac{\pi }{8}$ and $\frac{5\pi }{8}$  ·  $\frac{\pi }{8}$ only  ·  $\frac{\pi }{2}$ and $\frac{11\pi }{6}$  ·  $5$
   1. The maximum value of $3\sin x-4\cos x$ is .
   2. Solve $\tan 2x=1$ for $x\in [0,\pi [$. →
   3. Solve $\sin x+\sqrt{3}\cos x=1$ for $x\in [0,2\pi [$. →
   Show answer

   1: $5$ · 2: $\frac{\pi }{8}$ and $\frac{5\pi }{8}$ · 3: $\frac{\pi }{2}$ and $\frac{11\pi }{6}$

   • Step 1. $3\sin x-4\cos x=r\sin (x-\phi )$ with $r=\sqrt{9+16}=5$.
     Step 2. The largest value of $\sin$ is 1, so the maximum is 5.
   • Step 1. $2x=\frac{\pi }{4}+k\pi$.
     Step 2. $x=\frac{\pi }{8}+\frac{k\pi }{2}$: in $[0,\pi [$, $k=0,1$ give $\frac{\pi }{8}$, $\frac{5\pi }{8}$.
   • Step 1. $2\sin \left(x+\frac{\pi }{3}\right)=1$, so $\sin \left(x+\frac{\pi }{3}\right)=\frac{1}{2}$.
     Step 2. $x+\frac{\pi }{3}=\frac{\pi }{6}+2k\pi$ gives $x=-\frac{\pi }{6}\equiv \frac{11\pi }{6}$; $x+\frac{\pi }{3}=\frac{5\pi }{6}$ gives $x=\frac{\pi }{2}$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Solve $\sin 2x=\sin x$ for $x\in [0,\pi ]$.	a. $\frac{\sqrt{2}}{2}$
   2. Find the exact value of $\cos 15^{\circ }-\sin 15^{\circ }$.	b. $\frac{\pi }{3},\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{3}$
   3. Write $\sin x+\sqrt{3}\cos x$ in the form $r\sin (x+\phi )$.	c. $0,\frac{\pi }{3},\pi$
   4. Solve $2{\cos }^{2}x-\cos x=0$ for $x\in [0,2\pi [$.	d. $2\sin \left(x+\frac{\pi }{3}\right)$

   Show answer

   1 → c ($0,\frac{\pi }{3},\pi$) · 2 → a ($\frac{\sqrt{2}}{2}$) · 3 → d ($2\sin \left(x+\frac{\pi }{3}\right)$) · 4 → b ($\frac{\pi }{3},\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{3}$)

   • Step 1. $2\sin x\cos x-\sin x=\sin x(2\cos x-1)=0$.
     Step 2. $\sin x=0$: $0,\pi$. $\cos x=\frac{1}{2}$: $\frac{\pi }{3}$.
   • Step 1. $\cos x-\sin x=\sqrt{2}\cos (x+45^{\circ })$.
     Step 2. With $x=15^{\circ }$: $\sqrt{2}\cos 60^{\circ }=\frac{\sqrt{2}}{2}$.
   • Step 1. $r=\sqrt{1+3}=2$.
     Step 2. $\cos \phi =\frac{1}{2}$ and $\sin \phi =\frac{\sqrt{3}}{2}$: $\phi =\frac{\pi }{3}$.
   • Step 1. Factorise: $\cos x(2\cos x-1)=0$.
     Step 2. $\cos x=0$: $\frac{\pi }{2},\frac{3\pi }{2}$. $\cos x=\frac{1}{2}$: $\frac{\pi }{3},\frac{5\pi }{3}$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $\frac{24}{25}$  ·  $\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6}$  ·  $\frac{7\pi }{6},\frac{11\pi }{6}$ only  ·  $\tan x$
   1. Solve $2{\sin }^{2}x-\sin x-1=0$ for $x\in [0,2\pi [$. →
   2. Simplify $\frac{1-\cos 2x}{\sin 2x}$. →
   3. $a$ is acute with $\sin a=\frac{3}{5}$. Find $\sin 2a$. →
   Show answer

   1: $\frac{\pi }{2},\frac{7\pi }{6},\frac{11\pi }{6}$ · 2: $\tan x$ · 3: $\frac{24}{25}$

   • Step 1. $(2\sin x+1)(\sin x-1)=0$.
     Step 2. $\sin x=1$: $\frac{\pi }{2}$.
     Step 3. $\sin x=-\frac{1}{2}$: $\frac{7\pi }{6}$ and $\frac{11\pi }{6}$.
   • Step 1. $1-\cos 2x=2{\sin }^{2}x$ and $\sin 2x=2\sin x\cos x$.
     Step 2. $\frac{2{\sin }^{2}x}{2\sin x\cos x}=\frac{\sin x}{\cos x}$.
   • Step 1. $\cos a=\frac{4}{5}$ (acute).
     Step 2. $\sin 2a=2\times \frac{3}{5}\times \frac{4}{5}=\frac{24}{25}$.

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